1. Write the given lines in coordinate form

Given $\vec a = (1,2,1), \qquad \vec b = (2,7,3).$

So, $L_1: \vec r = (-1,2,1) + \lambda(1,2,1).$ Hence the parametric equations of $L_1$ are $x=-1+\lambda, \qquad y=2+2\lambda, \qquad z=1+\lambda.$

Also, $L_2: \vec r = (0,1,1) + \mu(2,7,3).$ Hence the parametric equations of $L_2$ are $x=2\mu, \qquad y=1+7\mu, \qquad z=1+3\mu.$

2. Find the point of intersection of $L_1$ and $L_2$

At the intersection, coordinates must be equal: $-1+\lambda = 2\mu \quad ...(1)$ $2+2\lambda = 1+7\mu \quad ...(2)$ $1+\lambda = 1+3\mu \quad ...(3)$

From (3), $\lambda = 3\mu.$

Substitute into (1): $-1+3\mu = 2\mu \implies \mu=1.$ Therefore, $\lambda=3.$

Check in (2): $2+2(3)=8, \qquad 1+7(1)=8,$ so it is consistent.

Thus the intersection point is $P = (-1+3,\; 2+2\cdot 3,\; 1+3) = (2,8,4).$

3. Find the direction vector of $L_3$

$L_3$ is parallel to $\vec a + \vec b$. So, $\vec a + \vec b = (1,2,1) + (2,7,3) = (3,9,4).$

Hence $L_3$ passes through $P=(2,8,4)$ and has direction vector $(3,9,4)$. Therefore, $L_3: \vec r = (2,8,4) + t(3,9,4).$

Its parametric form is $x=2+3t, \qquad y=8+9t, \qquad z=4+4t.$

4. Check which option lies on $L_3$

We test each option.

Option A: $(-1,-1,1)$

From $x=2+3t$, $-1=2+3t \implies t=-1.$ Then $y=8+9(-1)=-1$ correct, and $z=4+4(-1)=0 \neq 1.$ So A does not lie on $L_3$.

Option B: $(2,8,5)$

From $x=2+3t$, $2=2+3t \implies t=0.$ Then $z=4+4(0)=4 \neq 5.$ So B does not lie on $L_3$.

Option C: $(8,26,12)$

From $x=2+3t$, $8=2+3t \implies 3t=6 \implies t=2.$ Then $y=8+9(2)=26, \qquad z=4+4(2)=12.$ Both match. So C lies on $L_3$.

Option D: $(5,17,4)$

From $x=2+3t$, $5=2+3t \implies t=1.$ Then $y=8+9(1)=17$ correct, but $z=4+4(1)=8 \neq 4.$ So D does not lie on $L_3$.

5. Final answer

The required point is $\boxed{(8,26,12)}.$ So the correct option is C.